Advertisements
Advertisements
Question
Find expected value and variance of X for the following p.m.f.
| x | -2 | -1 | 0 | 1 | 2 |
| P(X) | 0.2 | 0.3 | 0.1 | 0.15 | 0.25 |
Advertisements
Solution 1
We construct the following table to calculate E (X) and V (X) :
| X = xi | pi =P [X = xi] | xi · pi | xi 2·pi = xi × xi·pi |
| -2 | 0.2 | -0.4 | 0.8 |
| -1 | 0.3 | -0.3 | 0.3 |
| 0 | 0.1 | 0 | 0 |
| 1 | 0.15 | 0.15 | 0.15 |
| 2 | 0.25 | 0.5 | 1 |
| Total | 1 | -0.05 | 2.25 |
From the table, Σxi · pi = -0.05 and Σxi2 · pi = 2.25
∴E (X) = Σxi · pi= -0.05
and V (X) = Σxi 2 ·pi - ( Σxi · pi)2
= 2.25 - (-0.05)2
= 2.25 - 0.0025 = 2.2475
Hence, E (X)= -0.05 and V (X) = 2.2475.
Solution 2
Expected value of X 5
= E(X) = \[\sum\limits_{i=1}^{5} x_i.\text{P}_i(x_i)\]
= (–2) x (0.2) + (–1) x (0.3) + 0 x (0.1) + 1 x (0.15) + 2 x (0.25)
= – 0.4 – 0.3 + 0 + 0.15 + 0.5
= – 0.05
E(X2) = \[\sum\limits_{i=1}^{5} x_i.\text{P}_i(x_i)\]
= (–2)2 x (0.2) + (–1)2 x (0.3) + 02 x (0.1) + 12 x (0.15) + 22 x (0.25)
= 0.8 + 0.3 + 0 + 0.15 + 1
= 2.25
∴ Variance of X
= Var(X)
= E(X2) – [E(X)]2
= 2.25 – (– 0.05)2
= 2.2475.
APPEARS IN
RELATED QUESTIONS
State if the following is not the probability mass function of a random variable. Give reasons for your answer.
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.1 | 0.5 | 0.2 | − 0.1 | 0.2 |
State if the following is not the probability mass function of a random variable. Give reasons for your answer.
| Y | −1 | 0 | 1 |
| P(Y) | 0.6 | 0.1 | 0.2 |
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the standard deviation of X.
The following is the p.d.f. of r.v. X:
f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise.
Find P (x < 1·5)
The following is the p.d.f. of r.v. X:
f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise.
P(x > 2)
It is known that error in measurement of reaction temperature (in 0° c) in a certain experiment is continuous r.v. given by
f (x) = `x^2 /3` , for –1 < x < 2 and = 0 otherwise
Verify whether f (x) is p.d.f. of r.v. X.
It is known that error in measurement of reaction temperature (in 0° c) in a certain experiment is continuous r.v. given by
f (x) = `x^2/3` , for –1 < x < 2 and = 0 otherwise
Find probability that X is negative
Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by f(x) = `1/5`, for 0 ≤ x ≤ 5 and = 0 otherwise.
Find the probability that the waiting time is more than 4 minutes.
Choose the correct option from the given alternative:
P.d.f. of a.c.r.v X is f (x) = 6x (1 − x), for 0 ≤ x ≤ 1 and = 0, otherwise (elsewhere)
If P (X < a) = P (X > a), then a = .....
Choose the correct option from the given alternative:
If the p.d.f of a.c.r.v. X is f (x) = 3 (1 − 2x2 ), for 0 < x < 1 and = 0, otherwise (elsewhere) then the c.d.f of X is F(x) =
Choose the correct option from the given alternative:
If a d.r.v. X takes values 0, 1, 2, 3, . . . which probability P (X = x) = k (x + 1)·5 −x , where k is a constant, then P (X = 0) =
Choose the correct option from the given alternative:
If p.m.f. of a d.r.v. X is P (X = x) = `((c_(x)^5 ))/2^5` , for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise If a = P (X ≤ 2) and b = P (X ≥ 3), then E (X ) =
Choose the correct option from the given alternative:
If p.m.f. of a d.r.v. X is P (X = x) = `x^2 /(n (n + 1))`, for x = 1, 2, 3, . . ., n and = 0, otherwise then E (X ) =
Solve the following :
Identify the random variable as either discrete or continuous in each of the following. Write down the range of it.
The person on the high protein diet is interested gain of weight in a week.
Solve the following problem :
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Identify the probability distribution of X and state the formula for p. m. f. of X.
The following is the c.d.f. of r.v. X
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| F(X) | 0.1 | 0.3 | 0.5 | 0.65 | 0.75 | 0.85 | 0.9 |
*1 |
P (–1 ≤ X ≤ 2)
The probability distribution of discrete r.v. X is as follows :
| x = x | 1 | 2 | 3 | 4 | 5 | 6 |
| P[x=x] | k | 2k | 3k | 4k | 5k | 6k |
(i) Determine the value of k.
(ii) Find P(X≤4), P(2<X< 4), P(X≥3).
Let X be amount of time for which a book is taken out of library by randomly selected student and suppose X has p.d.f
f (x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise.
Calculate: P(x≤1)
Let X be amount of time for which a book is taken out of library by randomly selected student and suppose X has p.d.f
f (x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise.
Calculate: P(0.5 ≤ x ≤ 1.5)
70% of the members favour and 30% oppose a proposal in a meeting. The random variable X takes the value 0 if a member opposes the proposal and the value 1 if a member is in favour. Find E(X) and Var(X).
Choose the correct alternative :
X: is number obtained on upper most face when a fair die….thrown then E(X) = _______.
Fill in the blank :
E(x) is considered to be _______ of the probability distribution of x.
State whether the following is True or False :
| x | – 2 | – 1 | 0 | 1 | 2 |
| P(X = x) | 0.2 | 0.3 | 0.15 | 0.25 | 0.1 |
If F(x) is c.d.f. of discrete r.v. X then F(–3) = 0
State whether the following is True or False :
If p.m.f. of discrete r.v. X is
| x | 0 | 1 | 2 |
| P(X = x) | q2 | 2pq | p2 |
then E(x) = 2p.
Solve the following problem :
The p.m.f. of a r.v.X is given by
`P(X = x) = {(((5),(x)) 1/2^5", ", x = 0", "1", "2", "3", "4", "5.),(0,"otherwise"):}`
Show that P(X ≤ 2) = P(X ≤ 3).
Solve the following problem :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
| x | – 1 | 0 | 1 |
| P(X = x) | `(1)/(5)` | `(2)/(5)` | `(2)/(5)` |
Solve the following problem :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
| X | 0 | 1 | 2 | 3 | 4 | 5 |
| P(X = x) | `(1)/(32)` | `(5)/(32)` | `(10)/(32)` | `(10)/(32)` | `(5)/(32)` | `(1)/(32)` |
Solve the following problem :
Let the p. m. f. of the r. v. X be
`"P"(x) = {((3 - x)/(10)", ","for" x = -1", "0", "1", "2.),(0,"otherwise".):}`
Calculate E(X) and Var(X).
Solve the following problem :
Let X∼B(n,p) If E(X) = 5 and Var(X) = 2.5, find n and p.
If the p.m.f. of a d.r.v. X is P(X = x) = `{{:(x/("n"("n" + 1))",", "for" x = 1"," 2"," 3"," .... "," "n"),(0",", "otherwise"):}`, then E(X) = ______
Find the expected value and variance of r.v. X whose p.m.f. is given below.
| X | 1 | 2 | 3 |
| P(X = x) | `1/5` | `2/5` | `2/5` |
Choose the correct alternative:
f(x) is c.d.f. of discete r.v. X whose distribution is
| xi | – 2 | – 1 | 0 | 1 | 2 |
| pi | 0.2 | 0.3 | 0.15 | 0.25 | 0.1 |
then F(– 3) = ______
The values of discrete r.v. are generally obtained by ______
The probability distribution of a discrete r.v.X is as follows.
| x | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X = x) | k | 2k | 3k | 4k | 5k | 6k |
Complete the following activity.
Solution: Since `sum"p"_"i"` = 1
k = `square`
The probability distribution of a discrete r.v.X is as follows.
| x | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X = x) | k | 2k | 3k | 4k | 5k | 6k |
Complete the following activity.
Solution: Since `sum"p"_"i"` = 1
P(X ≤ 4) = `square + square + square + square = square`
Using the following activity, find the expected value and variance of the r.v.X if its probability distribution is as follows.
| x | 1 | 2 | 3 |
| P(X = x) | `1/5` | `2/5` | `2/5` |
Solution: µ = E(X) = `sum_("i" = 1)^3 x_"i""p"_"i"`
E(X) = `square + square + square = square`
Var(X) = `"E"("X"^2) - {"E"("X")}^2`
= `sum"X"_"i"^2"P"_"i" - [sum"X"_"i""P"_"i"]^2`
= `square - square`
= `square`
The probability distribution of a discrete r.v. X is as follows:
| x | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X = x) | k | 2k | 3k | 4k | 5k | 6k |
- Determine the value of k.
- Find P(X ≤ 4)
- P(2 < X < 4)
- P(X ≥ 3)
The value of discrete r.v. is generally obtained by counting.
The p.m.f. of a random variable X is as follows:
P (X = 0) = 5k2, P(X = 1) = 1 – 4k, P(X = 2) = 1 – 2k and P(X = x) = 0 for any other value of X. Find k.
